Initial phase determination in a wave

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SUMMARY

The discussion centers on solving a wave equation represented as y=0.01cos(wt-kx)+0.02*6^(1/2)sin(wt-kx) with given parameters w=16s^-1 and k=6m^-1. The objective is to prove that the amplitude of the wave is 0.05 meters and the initial phase is -π/4. The user attempts to utilize phasor addition and trigonometric identities to derive the amplitude and phase but encounters difficulties in achieving the correct phase value. The solution requires a clear application of phasor addition techniques and trigonometric transformations.

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  • Understanding of wave equations and their components
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  • Familiarity with trigonometric identities and transformations
  • Basic principles of amplitude and phase in wave mechanics
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Homework Statement



I'm having trouble solving the following exercise:
Given the following wave equation:
y=0.01cos(wt-kx)+0.02*6^(1/2)sin(wt-kx)
prove that if w=16s^-1 and k=6m^-1 then the amplitude of the wave is 0.05 meters and the initial phase is -pi/4 how can I prove that? By phasor addiction probably but I can't go far that way. can someone show me how to do it?
Thank you

Homework Equations





The Attempt at a Solution


I converted sin(wt-kx) to cos(wt-kx-pi/2) so I can write the phasor corresponding to each cosine and attempt to add then an see what I get back, but what I get back I the equation I started with with the cos(wt-kx-pi/2) back as sin(wt-kx) by looking att the trignometric identity: acos(A)+bsin(B)=[tex]\sqrt{b^2+a^2}[/tex]*sin(A+C) where C is arctan([tex]\frac{b}{a}[/tex]) i could get the amplitude, but the phase I get is diffrent.
 
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can't you use trig to turn cos(A-B) into cosAcosB+sinAsinB, turn sin(A-B) to sinAcosB-cosAsinB
 

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